I have taken 2 semesters of Calculus at the University of Minnesota and 1 semester and the University of St. Thomas. I have to say, you are the greatest math instructor I have seen. I would not have passed my calculus courses without you, and the same goes for the thousands of other college students dependent upon your videos. Thank you.
So I've come back to this vid after a while for reference. I have pretty good intuition on finding limits for phi for whoever this may help: Essentially I think of phi as the maximum polar angle starting from the positive z-axis I would need in order to "sketch" whatever object I'm looking at. For the example of a sphere, you would only ever need to go up to pi radians in either direction to sketch a sphere. If we have, say, an eigth of a sphere in the first octant we would need pi/2 radians starting from the positive z-axis to sketch this object as the bottom of the eighth-sphere lies in the xy-plane. If we had a quarter or even half of a sphere resting on the xy-plane phi would still only be pi/2 (but angle theta would of course be different for these shapes). Same logic applies for shapes like cones but they're a little harder to figure out by inspection and you'll need to look at symmetry.
There’s a multitude of explanations on visualizing phi, but I want to share how I think about it. Think of the disk method from calc 2. When you made any simple circular shape, you would only need half of that shape to revolve it around the x or y axis. Phi is similar in that the angle range describes that half shape you’re initially gonna need, so that’s why it only goes from 0 to Pi: because after revolving it around the z-axis you would get a whole circle.
You were my salvation! You have no idea how desperate I was! RUclips doesn't have any videos about spheres coordenates in portuguese and my teachers can't teach, so you can imagine... Thank you soooooooo much!!!!
@soph6ia I think the relation x^2+y^2+z^2=p^2 is already accounted for in the dv formula he used (p^2*sin(phi)*dp*dphi*dtheta), so in including the function, he counted it twice and ended up with r^5/5 instead of r^3/3.
The (rho)^2 sin(phi) is actually the determinant of Jacobian matrix, which we must use when we are changing from one set of variables to other. I don't understand this stuff much, but what Patrick did is correct, the triple integral won't evaluate volume.
very nice!! i dont get how the hell you do this...i just stare at your video getting bits and pieces and then suddenly BAM i get the whole concept...as opposed to just getting bits and pieces :D Thanks alot you're seriously awesome!
I remember when I took this course. I got a B+. It was the most difficult math class I've ever had. I got A's in all other courses including differential equations with linear algebra!
+Jacob Krupp I was also so confused with that part and took me a while to figure it out. If you integrate w/ phi from zero to 2pi, you're integrating the whole circle + theta from zero to 2pi means you integrate the sphere twice. However, if you integrate phi from 0 to pi --> half sphere, then theta from zero to 2 pi --> full sphere!
6 лет назад+3
So basically think about the xyz axis in your head. Now on the right hand side a half sphere, that is the 0 to pi for the Phi part. And now, How would u make this a full circle? you would double it right. Thus you would put another one on the left side. Which is basically starting from 0 to 2pi for theta. No su have a full sphere. If you didn’t then you would integrate over 2 spheres as you would start with 1 whole spehere, then go from 0 to 360 again, basically doubling the sphere, which would create 2 spheres. Or maybe try and think of an animation in your head. That’s how I first learned it!! ;), from 0 to pi on z, y axis a half sphere is formed, then of x y, this sphere is lengthened from 0 to 2 pi create a full sphere. I hope this was of any assistance to anyone!! :D
Can we apply Fubini's theorem for any triple integral problem involving spherical coordinates? What if you are dealing with a sphere that is not centered at the origin? My professor never introduced it, so I didn't know that you could break up the integrals like that. I have an exam tomorrow and that trick would really help me out.
4 dislikes? WHY? All of these videos are fantastic. Helped me through AP Calc AB and BC in high school, a semester of multivariable calc in college, and right now I'm taking the calculus that comes after that (differential equations and stuff) and these videos STILL help me. We'll see what Calc 4 is like next semester hah Thanks Patrick!
The best way to think about it is if you draw a semi circle on a piece of paper. Then cut out your semicircle and spin it 360 degrees. When you do that, it makes one sphere without overlapping. If you cut out a circle from paper, you would make a sphere in half a turn but as a function is concerned the two halves would make an incomplete sphere because it would not be continuous.
Man, seriously your videos are a lifesaver for students everywhere. I am workin on my degree and I can definitely say you are helping SOOO much... I might ask them to present my degree in part to you,,, lol thanks so much
I don't know if you've watched the video response that showed it. After you were to integrate the theta from 0 to 2pi you basically get a disk of radius 1 and an infinitely small thickness. If you were to rotate that disk pi, you would get a sphere :)
GREAT VIDEO!....one question...i know Fubini's theorem is only applicable when the function is 'separable', but how do u know if a function is separable?
great tutorial to multivariable integration in spherical coordinates, however got the wrong answer.... the volume of a sphere is always (4piR^2)/3. You get this by simply putting a 1 in place of the p^2 you wrote. This is because a triple integral is actually 4d stuff and accommodates for the 3 dimensions. Is like taking a hypersphere and looki when its radius is fixed......
@Khanh Nguyen With a sphere, if you take a ray "p" and move it all the way from the positive z-axis to negative z-axis, you'll have created an angle of pi radians originating from the positive z-axis, which is called phi. If you then rotate that half circle shape 2pi in the xy plane (called angle theta) you'll have created a sphere. Phi ranges from 0 to pi, while theta ranges from 0 to 2pi. That is, crudely, the convention of spherical coordinates; hope that helps.
psynostic What I meant is that if you are trying to solve for volume, then you just integrate the differential of volume alone; V = int(1)*dV In this case dV is rho^2*sine(phi)*d(rho)*d(theta)*d(phi). And you can notice that after some algebra, the inside function can be simplified to rho^2, which is not 1!!!
Alexei Acevedo But the problem states to integrate the unit bell, which has magnitude of 1, thus x^2 + y^2 + z^2 = rho^2 = 1... ^2. So, (1)dV. "2:24 - 2.29"
psynostic Okay, since we are integrating a definite triple integral, we need 3 upper bounds and 3 lower bounds, which in this case will be defined by the unit sphere, however we will not obtain its volume, don't confuse. Extra information: triple integrals with inside functions other than 1 can us obtain the mass of an object defined by a unit sphere or square, or pyramid or even more complex shapes!
@andydrew129 When you're integrating in terms of any variable, treat the other variables like constants. So just like the (integral of 3x) = (3 * Integral of x), the same works if you take [sin(phi) * dphi * dtheta] out of the first integral (which is in terms of rho) out of that integral into the integral in terms of theta. The dtheta stays in that integral but the sin(phi)* dphi moves out into the last integral.
@soph6ia I agree with you that the volume is 4pi/3, but he is integrating a function f(x,y,z)=x^2+y^2+z^2 inside that sphere, so the answer of the integral won't be the volume of the sphere :)
i love u patrickjmt to the max!! you are such an amazing guy.! adore u! me and my friends always keep referring to you when we lost! :) keep it up! always be a great person on helping the entire world! truly love u!
Wow man! Thing thing looked so much more difficult from the beginning. I like that trick with breaking up the triple integral. My prof. likes to use “phi” as the x-y angle, and “theta” as the angle from the z axis. To me it makes more sense to do it your way.. even though you probably use whatever greek symbol you want.
Could you do an example using spherical coordinates to find the volume of a solid bounded by another solid? Say the volume outside of a cone but inside of a sphere.
THANKYOUTHANKYOUTHANKYOUTHANKYOUTHANKYOU!!!!! I don't know if you read comments on this video anymore, since idea s posted in 08, but you're teaching me Calc 3. 💛💙💜💚❤️
Im trying to apply this to a common calculation in astrophysics trying to find the standard value for Sin^3i to find the average inclination of orbit for binary star systems... Thank god for you taking time out to share your knowledge.. Now let me keep re watching this video until i can get it to fit into my application :-/ Cheers! ck
cheers, we are having alot of problems where our professor requires that we refer to a whole lot of theorems and definitions. I'd love to get these straightened out. Do you have any vid about using and understanding theorems like: gauss, green, stokes, fubini, Lagrange multipliers and the relationship between del, curl, grad and div?
I'm not seeing anything wrong with this like some are. I'm pretty certain Patrick is correct. He has the function (x2+y2+z2) to integrate, which is the same as p2. You have to multiply that p2 by p2 sin(phi), so you get p4 sin (phi) integrate p and you get p5/5, and eventually you get 4pi/5.
i have my calculus III exam tomorrow and this is saving my ass. I haven't been to class since October so I'm learning half of the course tonight. Ha. o well. Thanks!
Yes, I just watched the video. And I was looking if someone has already commented on this topic. I guess he should not include the p^2 in the integral. the correct form of integrating a triple integral is by just putting the boundaries without adding the function inside.
He did not only calculate volume, but also density. The triple integral would give the density of the unit ball and could tell us if it's a source or a sink. So, if this were a vector field, it would be considered a "source." The "row^2 sin (phi)" that's tacked on is called the Jacobian Transform. When the coordinate systems change, there's a Jacobian that must be added. For the simpler cylindrical coordinates, the Jacobian is just r, where r is the radius given in the form of (r, theta, z).
I have the most trouble when they give you a triple integral in cartesian coordinated and they give you the limits if integration. How do you convert the limits into spherical coordinates as well?
I actually think Ammaray yarammA's censored comment further down is correct. What we are calculating is not the Volume, it's the mass. If you try calculating the definite integral instead you'll get (4*pi*p^5)/5
Great video and very helpful. However, you did not say or show that rho squared sin phi was derived from the Jacobian which involves the determinant of x,y,z where... x = ρsinØcosϑ,y = ρsinØsinϑ,and z = ρcosØ. Thanks again for clarity.
Thanks for the lecture ! I'm thinking that the answer is 4/3 pi. The problem is that you took p'4 so it became p'5/5 and when you wrote 1 to p it generated that 1/5. However, when you take p'2 and integrate it, it becomes p'3/3 and it generates the 1/3 which is the correct. I guess that you explained it already but I wanted to write this explanation for who cannot understand what happened. Please warn me if I'm not right. Thanks for all of your lectures !
Hi, I think theres a mistake in the calculus, Im not sure where, but let see things this way : that integral should give us the volume of the unit sphere which is 4 pi/ 3. Your final result is 4 pi /5. Am I missing something ?
phi is measured from the positive z axis to point OP imagine O as your origin f(0,0,0) and P as some point between positive and negative z axis. make sure it is measured within your region paramaters
@Frank12387 You can hear Patrick say minus negative cosine at the part you are talking about. The opposite of a negative number is a positive number. or mathamatically, ((-cos(pi))-(-cos(0)) = -cos(pi) + cos(0)
Because of the Greek. They started pronouncing "Pi" wrong, because someone too many 'pee jokes.' π^2 = ππ. X) english.stackexchange.com/questions/11363/why-are-greek-letters-pronounced-incorrectly-in-scientific-english
Could you do another one where instead of what you are integrating being x^2+y^2+z^2 you integrate 1/(the square root of)x^2+y^2+z^2 with the sphere being x^2+y^2+z^2=25?
+theor14 No dude, it's not the volume. This is an integral over the volume of the unit sphere. It can be an electric field or something varying as x,y and z vary, and we want to count up the sum of these fields over the unit sphere. If you want to evaluate the volume of the unit sphere, you should write ∫∫∫dxdydz NOT ∫∫∫(x^2+y^2+z^2)dxdydz. I hope this helps..
***** What do you mean ? Like I said, this really isn't the calculations of the volume. This is just an integral over the volume. The function (x^2+y^2+z^2) could be the temperature of every point that's enclosed by this volume but this is not the value of the actual volume of the sphere. The volume of a sphere is ∫∫∫dxdydz where the limits of the integral create a sphere.
Hello. This is NOT setup for an integral into the 4th dimension. The hyperspace ("Volume") of a Hypersphere is 16*Pi/5 (r = 1). He should've known that p^2 = 1, this makes it easier. His mistake: He got = 4*Pi /5 Look for dimension 5: en.wikipedia.org/wiki/N-sphere en.wikipedia.org/wiki/Volume_of_an_n-ball#Balls_in_Lp_norms
***** Note that the volume of the sphere is not being determined here. If that were the case, the integrand would have had to have been equal to the value 1, since taking the triple integral of 1 is the same as taking the triple integral of dV, in which case we are simply summing an infinite amount of small volumes - so as to attain the entire volume. With this question rather, we are multiplying the the differential elements of volume by a function... Which has a very interesting geometrical interpretation.
how the hell did engineering students pass before the internet?? thanks a lot man.
Martin Maina when we graduate, we need to dedicate our degree to Patrick
Economics student too :X
because there was no internet
They read BOOKS!
@@opanpro9772 *Gasp* Books! you mean I had to spend hours looking into a bunch of clumped papers to find an answer? now thats impractical
What my teacher failed to explain in a week you've done in 6 minutes. Well played sir.
couldn't agree more
my professor conveniently forgot to mention that x^2 + y^2 + z^2 can be summarized by rho^2... minor details
Steve Wilson
Mine too, lol
A lot of these guys are good at math, but aren't good at explaining it to people that aren't already good at it.
Tony Tyrrell your teacher spent a week on this? You in the slow class bro
At least your teacher took a week, mine only spent a 1-hour lecture on it..... smh
I have taken 2 semesters of Calculus at the University of Minnesota and 1 semester and the University of St. Thomas. I have to say, you are the greatest math instructor I have seen. I would not have passed my calculus courses without you, and the same goes for the thousands of other college students dependent upon your videos. Thank you.
"Normally It's rho, I'm going to call it P" automatic like
Fr haha
Visualizing how to get the limits of integration for phi hurts my brain
So I've come back to this vid after a while for reference. I have pretty good intuition on finding limits for phi for whoever this may help:
Essentially I think of phi as the maximum polar angle starting from the positive z-axis I would need in order to "sketch" whatever object I'm looking at. For the example of a sphere, you would only ever need to go up to pi radians in either direction to sketch a sphere.
If we have, say, an eigth of a sphere in the first octant we would need pi/2 radians starting from the positive z-axis to sketch this object as the bottom of the eighth-sphere lies in the xy-plane. If we had a quarter or even half of a sphere resting on the xy-plane phi would still only be pi/2 (but angle theta would of course be different for these shapes).
Same logic applies for shapes like cones but they're a little harder to figure out by inspection and you'll need to look at symmetry.
Thanks, this helped me.
+Apoorv Patme Nice.
If you've got any other questions just ask! I got an A in calc 3 and feel like I have a pretty good grasp on the material.
@@TheArnoldification how do i contact you?
@@TheArnoldification Holy crap this is a really good way of looking at it. Thank you so much!!
if you could explain more examples of weird phi angles that would help
+Emilys Flaws yes its confusing thing of all
Emily Bäkken whats ur ig
There’s a multitude of explanations on visualizing phi, but I want to share how I think about it. Think of the disk method from calc 2. When you made any simple circular shape, you would only need half of that shape to revolve it around the x or y axis. Phi is similar in that the angle range describes that half shape you’re initially gonna need, so that’s why it only goes from 0 to Pi: because after revolving it around the z-axis you would get a whole circle.
You were my salvation! You have no idea how desperate I was! RUclips doesn't have any videos about spheres coordenates in portuguese and my teachers can't teach, so you can imagine... Thank you soooooooo much!!!!
@soph6ia I think the relation x^2+y^2+z^2=p^2 is already accounted for in the dv formula he used (p^2*sin(phi)*dp*dphi*dtheta), so in including the function, he counted it twice and ended up with r^5/5 instead of r^3/3.
funny how this vid was made ~11 years ago and Its still being used by students like me
i could remake the exact same video today so that everyone feels better about it :)
Math doesn't really change much. This stuff was discovered hundreds of years ago. Sometimes notation changes over time though
Make it 12
Make it 13
make it 15@@homerosu353
The (rho)^2 sin(phi) is actually the determinant of Jacobian matrix, which we must use when we are changing from one set of variables to other. I don't understand this stuff much, but what Patrick did is correct, the triple integral won't evaluate volume.
wonderful!
i know! i was there. that is why i made this video!
very nice!! i dont get how the hell you do this...i just stare at your video getting bits and pieces and then suddenly BAM i get the whole concept...as opposed to just getting bits and pieces :D
Thanks alot you're seriously awesome!
WOW totally forgot that you can break up the integrals like that.. Thank you SOOO much, saved me much grief
I remember when I took this course. I got a B+. It was the most difficult math class I've ever had.
I got A's in all other courses including differential equations with linear algebra!
@Shadoweprinz yes, i wish more people got involved - i would be fine with video responses
thanks for the kind words ;)
@iamarealpersonsrsly glad the class turned out well for you!
I'm just a little confused on the phi part where you go from zero to Pi
+Jacob Krupp I was also so confused with that part and took me a while to figure it out. If you integrate w/ phi from zero to 2pi, you're integrating the whole circle + theta from zero to 2pi means you integrate the sphere twice. However, if you integrate phi from 0 to pi --> half sphere, then theta from zero to 2 pi --> full sphere!
So basically think about the xyz axis in your head. Now on the right hand side a half sphere, that is the 0 to pi for the Phi part. And now, How would u make this a full circle? you would double it right. Thus you would put another one on the left side. Which is basically starting from 0 to 2pi for theta. No su have a full sphere. If you didn’t then you would integrate over 2 spheres as you would start with 1 whole spehere, then go from 0 to 360 again, basically doubling the sphere, which would create 2 spheres. Or maybe try and think of an animation in your head. That’s how I first learned it!! ;), from 0 to pi on z, y axis a half sphere is formed, then of x y, this sphere is lengthened from 0 to 2 pi create a full sphere. I hope this was of any assistance to anyone!! :D
Goku17yen wow thank you so much I visualized that animation clearly in my head. Here's hoping I do well my final
No prob man, happy to help 😃
@tadm123 it depends on the compound to be analyzed. for xample: if you have a compound bounded by
4 =< X^2 + Y^2
Can we apply Fubini's theorem for any triple integral problem involving spherical coordinates? What if you are dealing with a sphere that is not centered at the origin? My professor never introduced it, so I didn't know that you could break up the integrals like that. I have an exam tomorrow and that trick would really help me out.
I don't often leave comments, but I had to here. Your tutorials are amazing, truly truly helpful. Bless you sir.
bro wtf this so good thank you man.
4 dislikes? WHY? All of these videos are fantastic. Helped me through AP Calc AB and BC in high school, a semester of multivariable calc in college, and right now I'm taking the calculus that comes after that (differential equations and stuff) and these videos STILL help me. We'll see what Calc 4 is like next semester hah
Thanks Patrick!
@sdjkl452 thanks! glad u like it!
The best way to think about it is if you draw a semi circle on a piece of paper. Then cut out your semicircle and spin it 360 degrees. When you do that, it makes one sphere without overlapping. If you cut out a circle from paper, you would make a sphere in half a turn but as a function is concerned the two halves would make an incomplete sphere because it would not be continuous.
Man, seriously your videos are a lifesaver for students everywhere. I am workin on my degree and I can definitely say you are helping SOOO much... I might ask them to present my degree in part to you,,, lol thanks so much
I don't know how, but I understood almost everything you said perfectly. Thanks a lot dude!!!
I don't know if you've watched the video response that showed it. After you were to integrate the theta from 0 to 2pi you basically get a disk of radius 1 and an infinitely small thickness. If you were to rotate that disk pi, you would get a sphere :)
GREAT VIDEO!....one question...i know Fubini's theorem is only applicable when the function is 'separable', but how do u know if a function is separable?
great tutorial to multivariable integration in spherical coordinates, however got the wrong answer.... the volume of a sphere is always (4piR^2)/3. You get this by simply putting a 1 in place of the p^2 you wrote. This is because a triple integral is actually 4d stuff and accommodates for the 3 dimensions. Is like taking a hypersphere and looki when its radius is fixed......
I saw your explanation and I liked so much, i'm brazilian n i studin' civil engineering n 4 us it's very important. Congratulations, u did a nice job
@Khanh Nguyen With a sphere, if you take a ray "p" and move it all the way from the positive z-axis to negative z-axis, you'll have created an angle of pi radians originating from the positive z-axis, which is called phi. If you then rotate that half circle shape 2pi in the xy plane (called angle theta) you'll have created a sphere. Phi ranges from 0 to pi, while theta ranges from 0 to 2pi. That is, crudely, the convention of spherical coordinates; hope that helps.
Hey, I have been trying to find your Cylindrical Coordinates video but cant seem to find one. Do you not have one for that topic? Thanks
@Matmo11 only in certain cases
wow thats a really helpful use of fubini,
does that trick only work in spherical coordinates pat?
I was trying to explain this to a 15 year old. He didn't understand it. :D
Thanks for the awesome videos Patrick!
Isnt the volume of a sphere of radius 1 is equal to 4pi/3?
Yes! But we are not integrating volume, unless our inside function is 1.
Alexei Acevedo Why, do you say, is the meaning of an extra Rho-Square?
psynostic What I meant is that if you are trying to solve for volume, then you just integrate the differential of volume alone; V = int(1)*dV In this case dV is rho^2*sine(phi)*d(rho)*d(theta)*d(phi). And you can notice that after some algebra, the inside function can be simplified to rho^2, which is not 1!!!
Alexei Acevedo But the problem states to integrate the unit bell, which has magnitude of 1, thus x^2 + y^2 + z^2 = rho^2 = 1... ^2. So, (1)dV. "2:24 - 2.29"
psynostic Okay, since we are integrating a definite triple integral, we need 3 upper bounds and 3 lower bounds, which in this case will be defined by the unit sphere, however we will not obtain its volume, don't confuse.
Extra information: triple integrals with inside functions other than 1 can us obtain the mass of an object defined by a unit sphere or square, or pyramid or even more complex shapes!
@andydrew129 When you're integrating in terms of any variable, treat the other variables like constants. So just like the (integral of 3x) = (3 * Integral of x), the same works if you take [sin(phi) * dphi * dtheta] out of the first integral (which is in terms of rho) out of that integral into the integral in terms of theta. The dtheta stays in that integral but the sin(phi)* dphi moves out into the last integral.
@soph6ia I agree with you that the volume is 4pi/3, but he is integrating a function f(x,y,z)=x^2+y^2+z^2 inside that sphere, so the answer of the integral won't be the volume of the sphere :)
best calculus videos on youtube.. really helpful
please do an example using cylindrical coordinates and thanks for all your help!
great comment! thanks!!
Will you do some examples with cylindrical coordinates?
i had also never seen that trick for breaking up integral evaluation--nice
i love u patrickjmt to the max!! you are such an amazing guy.! adore u! me and my friends always keep referring to you when we lost! :) keep it up! always be a great person on helping the entire world! truly love u!
Excelent video, after seeing this video ,problems were easy as plucking a flower,
you change mylife THANKYOU!!!!
Wow man! Thing thing looked so much more difficult from the beginning. I like that trick with breaking up the triple integral. My prof. likes to use “phi” as the x-y angle, and “theta” as the angle from the z axis. To me it makes more sense to do it your way.. even though you probably use whatever greek symbol you want.
im looking the "phi part" all spanish side of the web...now i found it, in english, but found it! Thanks man!
well, just play w/ the algebra. u need it to be the product of a fuction of x and a function of y:
f(x)g(y)
Thank you very much,please dont stop with this.. u are great !!
Could you do an example using spherical coordinates to find the volume of a solid bounded by another solid? Say the volume outside of a cone but inside of a sphere.
@Frank12387 Not quite, you can hear Patrick say, minus a negative cosine as he makes the plus. The opposite of a negative number is a positive number.
T_T do a harder one, please! the question on my exam will be harder than this.
@omarofuae you are a good soul : )
i hope you post more of these spherical. probably much harder ones. thanks! :D
Not that anyone is surprised that Patrick is right again!!
THANKYOUTHANKYOUTHANKYOUTHANKYOUTHANKYOU!!!!! I don't know if you read comments on this video anymore, since idea s posted in 08, but you're teaching me Calc 3. 💛💙💜💚❤️
my summer term instructor sucks , he didn’t explain why Phi range from 0 to Pi,
Im trying to apply this to a common calculation in astrophysics trying to find the standard value for Sin^3i to find the average inclination of orbit for binary star systems... Thank god for you taking time out to share your knowledge..
Now let me keep re watching this video until i can get it to fit into my application :-/
Cheers!
ck
cheers, we are having alot of problems where our professor requires that we refer to a whole lot of theorems and definitions. I'd love to get these straightened out. Do you have any vid about using and understanding theorems like: gauss, green, stokes, fubini, Lagrange multipliers and the relationship between del, curl, grad and div?
Wow man, I think you would make a great calculus prof! I can understand your way of teaching very well.
PLEASE PLEASE make a video of changing the order of integration in triple integrals
dude i think i just passed my math exam....you rock!
I'm not seeing anything wrong with this like some are. I'm pretty certain Patrick is correct. He has the function (x2+y2+z2) to integrate, which is the same as p2. You have to multiply that p2 by p2 sin(phi), so you get p4 sin (phi) integrate p and you get p5/5, and eventually you get 4pi/5.
i have my calculus III exam tomorrow and this is saving my ass. I haven't been to class since October so I'm learning half of the course tonight. Ha. o well. Thanks!
You are a fucking legend!
Thanks from undergraduate engineer.
you are very welcome :)
I'm confused about the phi Part, Patrick can you help?
hey patrick, can you do a more difficult spherical coord problem?
Has anyone commented that the volume of a sphere is not 4pi/5? It's 4pi/3 because you dont put the extra rho ^2 in at the start like he did.
Right. We actually integrate 1 as opposed to x^2+y^2 + z^2
I did notice. It was a head scratcher had me grab paper and pencil to work it out.
Yes, I just watched the video. And I was looking if someone has already commented on this topic. I guess he should not include the p^2 in the integral. the correct form of integrating a triple integral is by just putting the boundaries without adding the function inside.
The example is correct. Triple integrals don't evaluate volume but I guess you could say properties of volume within a given region.
He did not only calculate volume, but also density. The triple integral would give the density of the unit ball and could tell us if it's a source or a sink. So, if this were a vector field, it would be considered a "source."
The "row^2 sin (phi)" that's tacked on is called the Jacobian Transform. When the coordinate systems change, there's a Jacobian that must be added. For the simpler cylindrical coordinates, the Jacobian is just r, where r is the radius given in the form of (r, theta, z).
I have the most trouble when they give you a triple integral in cartesian coordinated and they give you the limits if integration. How do you convert the limits into spherical coordinates as well?
I actually think Ammaray yarammA's censored comment further down is correct. What we are calculating is not the Volume, it's the mass. If you try calculating the definite integral instead you'll get (4*pi*p^5)/5
Great video and very helpful. However, you did not say or show that rho squared sin phi was derived from the Jacobian which involves the determinant of x,y,z where...
x = ρsinØcosϑ,y = ρsinØsinϑ,and z = ρcosØ.
Thanks again for clarity.
Oh my gah!! You can split the integrals??? My life would've been so much easier had I known this earlier. =/
Thanks so much for this!!
can you split up all types of integrals or only spherical ones?
learning has become easy thanks you guys
dude that is amazing, so clear.
i just got some new information from you
thanks
I always like your video before watching it because I know it's gonna help me:)
In which cases can you break the integral like that?
could you do examples of triple integrals in cylindrical coordinates?
this subject definitely need more examples, with this example alone doesn't give me the confidence to work with spherical coordinates
Thanks for the lecture ! I'm thinking that the answer is 4/3 pi. The problem is that you took p'4 so it became p'5/5 and when you wrote 1 to p it generated that 1/5. However, when you take p'2 and integrate it, it becomes p'3/3 and it generates the 1/3 which is the correct. I guess that you explained it already but I wanted to write this explanation for who cannot understand what happened. Please warn me if I'm not right. Thanks for all of your lectures !
patrick you are my hero and my saviour
Hi, I think theres a mistake in the calculus, Im not sure where, but let see things this way :
that integral should give us the volume of the unit sphere which is 4 pi/ 3. Your final result is 4 pi /5. Am I missing something ?
you are good. I have been wondering about triple integrals
phi is measured from the positive z axis to point OP
imagine O as your origin f(0,0,0) and P as some point between positive and negative z axis.
make sure it is measured within your region paramaters
@Frank12387 You can hear Patrick say minus negative cosine at the part you are talking about. The opposite of a negative number is a positive number.
or mathamatically, ((-cos(pi))-(-cos(0)) = -cos(pi) + cos(0)
Is the only case when you can break them up is when the rest of the equation simply acts as a constant?
Why do you pronounce pi as pie, but you pronounce phi as phee?
Because of the Greek. They started pronouncing "Pi" wrong, because someone too many 'pee jokes.' π^2 = ππ. X)
english.stackexchange.com/questions/11363/why-are-greek-letters-pronounced-incorrectly-in-scientific-english
I'm eating pi(e), right now....Lemon meringue ...and it's integrated with a very soft crust with lots of cinnamon .
very nice explanation! Thank you
Wow,thanks!! Continue making helpful videos like this
HI, thanks for the explanation. Just one question. Can you explain bit more about how you got the limit of phi from 0 to Pi please?
Thank you
Could you do another one where instead of what you are integrating being x^2+y^2+z^2 you integrate 1/(the square root of)x^2+y^2+z^2 with the sphere being x^2+y^2+z^2=25?
So Is it ok to say that PHI is always going to be half of theta?
Why don't you put this in your Calc 3 playlist?
Why isn't the volume 4pi/3 ?
+theor14 No dude, it's not the volume. This is an integral over the volume of the unit sphere. It can be an electric field or something varying as x,y and z vary, and we want to count up the sum of these fields over the unit sphere. If you want to evaluate the volume of the unit sphere, you should write ∫∫∫dxdydz NOT ∫∫∫(x^2+y^2+z^2)dxdydz. I hope this helps..
+Abdulrahman Mahdaly thank you.
DON'T let them CONFUSE you! The answer's WRONG. They looked for VOLUME, and we BOTH knew the answer was 4pi/3. I talk about this above.
***** What do you mean ? Like I said, this really isn't the calculations of the volume. This is just an integral over the volume. The function (x^2+y^2+z^2) could be the temperature of every point that's enclosed by this volume but this is not the value of the actual volume of the sphere. The volume of a sphere is ∫∫∫dxdydz where the limits of the integral create a sphere.
Hello. This is NOT setup for an integral into the 4th dimension. The hyperspace ("Volume") of a Hypersphere is 16*Pi/5 (r = 1). He should've known that p^2 = 1, this makes it easier. His mistake: He got = 4*Pi /5
Look for dimension 5:
en.wikipedia.org/wiki/N-sphere
en.wikipedia.org/wiki/Volume_of_an_n-ball#Balls_in_Lp_norms
***** Note that the volume of the sphere is not being determined here. If that were the case, the integrand would have had to have been equal to the value 1, since taking the triple integral of 1 is the same as taking the triple integral of dV, in which case we are simply summing an infinite amount of small volumes - so as to attain the entire volume. With this question rather, we are multiplying the the differential elements of volume by a function... Which has a very interesting geometrical interpretation.
how you do the that trick of integration for cylindrical coordinate as well?